Question Derive the following formula for the sum of triangu

Question: Derive the following formula for the sum of triangular numbers. attributed to the Hindu mathematician Aryabhata (circa 500 AD): t1 + t2 + t3 + ... + tn = (n + 1)(n + 2)/6 n >= 1 [Hint: Group the terms on the left-hand side in pairs. noting the identity, tk-1 + tk = k2]

Consider the following cubes:

1^3 = (1+0)^3 = 1^3
2^3 = (1+1)^3 = 1^3 +3(1